Lesson 9-2

Example 1
Graph r = -cos .

Make a table of values. Round the values of r to the nearest tenth. Graph the ordered pairs and connect them with smooth curve.

	-cos 	(r, )	

	Notice that the ordered pairs obtained when 180° ≤  ≤ 360° represent the same points as the ordered pairs obtained when 0° ≤  ≤ 180°.	
0°
30°
45°
60°
90°
120°
135°
150°
180°
210°
225°
240°
270°
300°
315°
330°
360°	-1
-0.9
-0.7
-0.5
0
0.5
0.7
0.9
1
0.9
0.7
0.5
0
-0.5
-0.7
-0.9
-1	(-1, 0°)
(-0.9, 30°)
(-0.7, 45°)
(-0.5, 60°)
(0, 90°)
(0.5, 120°)
(0.7, 135°)
(0.9, 150°)
(1, 180°)
(0.9, 210°)
(0.7, 225°)
(0.5, 240°)
(0, 270°)
(-0.5, 300°)
(-0.7, 315°)
(-0.9, 330°)
(-1, 360°)		

Example 2
Graph each polar equation.

a.	r = 2 + 3 sin 

	2 + 3 sin 	(r, )		
0	2	(2, 0)		
	3.5	3.5, 		
	4.6	4.6, 		
	5	5, 		
	4.6	4.6, 		
	3.5	3.5, 		
	2	(2, )		
	0.5	0.5, 		
	-0.6	-0.6, 		
	-1	-1, 		
	-0.6	-0.6, 		
	0.5	0.5, 		
2	2	(2, 2)		

b.	r = -2 cos 3

	-2 cos 3	(r, )		
0	-2	(-2, 0)		
	0	0, 		
	2	2, 		
	0	0, 		
	-2	-2, 		
	0	0, 		
	2	(2, )		
	0	0, 		
	-2	-2, 		
	0	0, 		
	2	2, 		
	0	0, 		
2	-2	(-2, 2)		

Example 3
AUDIO TECHNOLOGY  Another type of microphone is called a hypercardiod
microphone. The polar pattern a hypercardiod microphone is modeled by the polar equation r = 0.25 + 0.75 cos .

a.	Sketch a graph of the polar pattern.

b.	Describe what the polar pattern tells you about the microphone.

a.		0.25 + 0.75 cos 	(r, )		
	0°
30°
60°
90°
120°
150°
180°
210°
240°
270°
300°
330°
360°	1
0.9
0.6
0.3
-0.1
-0.4
-0.5
-0.4
-0.1
0.3
0.6
0.9
1	(1, 0°)
(0.9, 30°)
(0.6, 60°)
(0.3, 90°)
(-0.1, 120°)
(-0.4, 150°)
(-0.5, 180°)
(-0.4, 210°)
(-0.1, 240°)
(0.3, 270°)
(0.6, 300°)
(0.9, 330°)
(1, 360°)		

	The type of curve is called a limaçon.

b.	The polar pattern indicates that the microphone will pick up very loud sounds that are behind it, but it will also pick up soft sounds that are either directly in front of it or directly behind it.

Example 4
Graph the system of polar equations. Solve the system using algebra and trigonometry, and compare the solutions to those on your graph.
	r = 4 + 4 cos 
	r = 3 + 2 cos 

To solve the system of equations,	
substitute 4 + 4 cos   for r in the
second equation.

	4 + 4 cos 	= 3 + 2 cos 
	2 cos 	= -1
	cos 	= -
		=  or  = 

Substituting each angle into either of the original equations gives r = 2, so the solutions of the system are 2,  and 2, . Tracing the curves shows that these solutions correspond with the intersection points of the graphs.